# Circuit- Power absorbed

Discussion in 'Homework Help' started by J.live, Sep 12, 2010.

1. ### J.live Thread Starter Member

Sep 10, 2010
35
0
Find the power absorbed by Rx

Given Rx = 5 ohms

Attempt:

-20 + 1.5i + 5i=0 (5i is the total of the resistors in series and parallel on the right)

i= 6.5

p= (i)^2 (R) = (6.5)^2 * 5 = 211. 25 W ?

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2. ### Georacer Moderator

Nov 25, 2009
5,151
1,266
Didn't we go over an entire thread previously to calculate Req? Use it in your calculations to find the total resistance of the circuit. Only then try find the current flowing through the voltage source.

3. ### J.live Thread Starter Member

Sep 10, 2010
35
0
I did use Req to find the total resistance.

Req = 5 ohms.

So I used it as 5i in the equation.

-20+1.5i+ 5i=0

4. ### Ghar Active Member

Mar 8, 2010
655
73
Where is Rx in that equation?

It's correct if Rx didn't exist.

5. ### J.live Thread Starter Member

Sep 10, 2010
35
0
Oh, shoot. Sorry.

It should be

-20+1.5i+ 5i + 5i = 0

i= 1.7 A

p= (1.7)^2 (5)= 14.45 W ?

6. ### Georacer Moderator

Nov 25, 2009
5,151
1,266
That's not quite correct...
Let's take the facts from the beginning: Your exercise stated that for this part of the problem we will assume that Rx=5 Ohms. This information needs to be used.

So the total resistance is (1.5+(Rx||Req)) Ohms = (1.5+(5||5)) Ohms, wich is not 6.5 as you wrote above.

You also solved the equation wrong:
but don't linger on it since this insn't the equation you want either.

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7. ### Ghar Active Member

Mar 8, 2010
655
73
Req is in parallel with Rx so that equation is a bit wrong.

That 1.7, if correct, would be the current in the voltage source. You need the current in Rx to calculate the power in Rx.

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8. ### J.live Thread Starter Member

Sep 10, 2010
35
0
Okay I get what I was doing wrong. Since RxllReq I cant just add them up like series

Rtotal= 4

-20 + 4i=0

i= 5

p= (5)^2*5 = 125

Wait, I need to find a separate current for Rx?

If so can I use the current division rule?

Last edited: Sep 12, 2010
9. ### Ghar Active Member

Mar 8, 2010
655
73
Yes, current division works.

You have the current in the source, which is also the current in the 1.5 ohm resistor.

It then splits between Rx and Req.

10. ### J.live Thread Starter Member

Sep 10, 2010
35
0
Okay

i= 10x 5/ (5+5)

i = 5 A

p= (5)^2 (5) = 125 W.

It still derives to the same answer.

I suppose this is a sheer coincident.

11. ### Georacer Moderator

Nov 25, 2009
5,151
1,266
Or it could be a mistake...

You already found that the current through the source is 5 A. That means that the current through the 1.5 Ohm resistance is the same, wich is then divided between Rx and Req. The current through Rx is $Ix=I_{total} \cdot \frac{R_{eq}}{R_x+R_{eq}}=2.5 A$.

Wich makes the power as much as $P=I^2 \cdot R_x=2.5^2 \cdot 5 W=31.25 W$

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12. ### Ghar Active Member

Mar 8, 2010
655
73
No, it's a mistake

That would imply that the all the current is going through Rx and none through Req.
Rx = Req, so how can one take all the current?

I'm not sure what equation you just wrote there.
Where did that 10 come from?

13. ### J.live Thread Starter Member

Sep 10, 2010
35
0
Oh wow. Now i am embarrassed of myself.

Sorry guys. Idk why I can't look out for these stupid mistakes.